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A local bank that has 15 branches uses a two-digit code to [#permalink]
27 Jul 2010, 11:25

1

This post was BOOKMARKED

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Difficulty:

25% (medium)

Question Stats:

66% (01:59) correct
34% (00:55) wrong based on 131 sessions

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

Re: A local bank that has 15 branches uses a two-digit code to [#permalink]
27 Mar 2013, 23:05

1

This post received KUDOS

Let the required number of digits be n. Considering the given conditions, i)The first digit can be filled up in n ways. ii)The second digit can be filled up in n ways too. So we will get n * n = n^2 numbers.

n^2 \geq 15

=>n \geq 4 [since n is an integer.]

So,option B will be the correct answer. ------------------------------------------- Please press KUDOS if you like my post. _________________

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

Choices A 3

B 4

C 5

D 6

E 7

Consider the code XY. If there are n digits available then X can take n values and Y can also take n values, thus from n digits we can form n^2 different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).

We want # of codes possible from n digit to be at least 15 --> n^2\geq{15} --> n\geq4, hence min 4 digits are required.

I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain? _________________

I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?

It's always good to test theoretical thoughts on practice:

How many codes can be formed using 2 digits (n=2), 0 and 1.:

00; 01; 10; 11.

4=2^2.

Or consider the following: how many 2-digit codes can be formed out of 10 digits (0, 1, 2, 3, ..., 9)? 00; 01; 02; ... 99.

I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?

n * (n-1) is to say that 99 will not be chosen. To choose 99 once we are saying n*n should be the combo _________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Re: Permutation question [#permalink]
31 Jan 2011, 16:58

Bunuel wrote:

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

Choices A 3

B 4

C 5

D 6

E 7

Consider the code XY. If there are n digits available then X can take n values and Y can also take n values, thus from n digits we can form n^2 different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).

We want # of codes possible from n digit to be at least 15 --> n^2\geq{15} --> n\geq4, hence min 4 digits are required.

Answer: B.

Hope it's clear.

Actually it could be A. B/c think of these arrangements for the 15 codes.

00, 01, 10, 02, 20, 03, 30, 11, 21, 12, 31, 13, 22, 23, 32 and 33. We have 16 arrangements so minimum # of different integers used can be 3.

What do you think? _________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

Re: Permutation question [#permalink]
31 Jan 2011, 17:03

Expert's post

mariyea wrote:

Bunuel wrote:

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

Choices A 3

B 4

C 5

D 6

E 7

Consider the code XY. If there are n digits available then X can take n values and Y can also take n values, thus from n digits we can form n^2 different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).

We want # of codes possible from n digit to be at least 15 --> n^2\geq{15} --> n\geq4, hence min 4 digits are required.

Answer: B.

Hope it's clear.

Actually it could be A. B/c think of these arrangements for the 15 codes.

00, 01, 10, 02, 20, 03, 30, 11, 21, 12, 31, 13, 22, 23, 32 and 33. We have 16 arrangements so minimum # of different integers used can be 3.

Re: Permutation question [#permalink]
02 Feb 2011, 09:22

Starting with choice A, 3 * 3 = 9 options are possible to code 15 branches. Not suff. Using 4 in choice B, 4 * 4 = 16 options are possible. We need the fewest. So B.

A local bank that has 15 branches uses a two-digit code to r [#permalink]
26 Mar 2013, 09:19

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer cannot be used for both digits of a code, are considered as two separate codes. What is the fewest number of different integers required for the 15 codes? Choices A 3

B 4

C 5

D 6

E 7 _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: A local bank that has 15 branches uses a two-digit code to r [#permalink]
26 Mar 2013, 09:35

N= number of integers We have Noptions for the first one, N-1 options for the second. So a total of N(N-1) combinations, and we want that N(N-1)=15. Now or you plug in the different options, or you solve N^2-N-15=0; the first way seems faster, so lets try with A) 3 : 3*2=6 No B)4*3=12 No again C)5*4=20 YES

C _________________

It is beyond a doubt that all our knowledge that begins with experience.

Then I highlight the possible combinations (not considering the numbers with repeated integers) For example with 1 integer there are nor any number possible, with 2 integers, 2 possible numbers, with 3 integers , 6 possible numbers, with 4 integers, 12 possible numbers, and with 5 ... 20 . Correct Answer C

I know it´s not the finest answer, I guess it should be explained with combinatory.

Re: A local bank that has 15 branches uses a two-digit code to r [#permalink]
27 Mar 2013, 03:48

Expert's post

skamal7 wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer cannot be used for both digits of a code, are considered as two separate codes. What is the fewest number of different integers required for the 15 codes? Choices A 3

B 4

C 5

D 6

E 7

Merging similar topics. Please refer to the discussion above.

Re: A local bank that has 15 branches uses a two-digit code to [#permalink]
16 Aug 2014, 01:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: A local bank that has 15 branches uses a two-digit code to [#permalink]
19 Aug 2014, 07:24

More easy solution. Think logically.

Pick any two integer.

Integers: 1 & 2

Code: 11, 12, 21, 22 = 4 Codes

Add one more integer: 3

13, 31, 33, 23, 32 = 5 Codes

Add one more integer: 4

44, 14, 41, 24, 42, 34, 43 = 7 Codes

Total = 16 Codes. Enough. Answer: B

2 integers create 4 codes. we need 15 codes.

zest4mba wrote:

A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?

A. 3 B. 4 C. 5 D. 6 E. 7

_________________

If you like my post give me kudos.

Arindam Sur Researcher, Academian

gmatclubot

Re: A local bank that has 15 branches uses a two-digit code to
[#permalink]
19 Aug 2014, 07:24